Lemma 11.4.3. Let $V$ be a $k$ vector space. Let $K$ be a central $k$-algebra which is a skew field. Let $W \subset V \otimes _ k K$ be a two-sided $K$-sub vector space. Then $W$ is generated as a left $K$-vector space by $W \cap (V \otimes 1)$.

**Proof.**
Let $V' \subset V$ be the $k$-sub vector space generated by $v \in V$ such that $v \otimes 1 \in W$. Then $V' \otimes _ k K \subset W$ and we have

If $\overline{v} \in V/V'$ is a nonzero vector such that $\overline{v} \otimes 1$ is contained in $W/(V' \otimes _ k K)$, then we see that $v \otimes 1 \in W$ where $v \in V$ lifts $\overline{v}$. This contradicts our construction of $V'$. Hence we may replace $V$ by $V/V'$ and $W$ by $W/(V' \otimes _ k K)$ and it suffices to prove that $W \cap (V \otimes 1)$ is nonzero if $W$ is nonzero.

To see this let $w \in W$ be a nonzero element which can be written as $w = \sum _{i = 1, \ldots , n} v_ i \otimes k_ i$ with $n$ minimal. We may right multiply with $k_1^{-1}$ and assume that $k_1 = 1$. If $n = 1$, then we win because $v_1 \otimes 1 \in W$. If $n > 1$, then we see that for any $c \in K$

and hence $c k_ i - k_ i c = 0$ by minimality of $n$. This implies that $k_ i$ is in the center of $K$ which is $k$ by assumption. Hence $w = (v_1 + \sum k_ i v_ i) \otimes 1$ contradicting the minimality of $n$. $\square$

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